Question

A simple pendulum is released from A as shown. If m and l represent the mass of the bob and length of the pendulum, the gain in kinetic energy at B is :

• A. $\frac { \mathrm { mg } \ell } { 2 }$
• B. $\frac { \mathrm { mg } \ell } { \sqrt { 2 } }$
• C. $\frac { \sqrt { 3 } } { 2 } \mathrm { mg } \ell$
• D. $\frac { 2 } { \sqrt { 3 } } \mathrm { mg } \ell$

Answer 1

Answer: (C)

$\frac { \sqrt { 3 } } { 2 } \mathrm { mg } \ell$

Answer 2

DK + DU = 0 ; (K – 0) + (– mgh) = 0

K = mgh = mg (l – d), and

d = l(1 – cos 30⁰)

\ K = mg l cos 30⁰ = $\sqrt { 3 } / 2$mgl